MYP · 09 · extended response
Identifying the missing fundamental on a string
Waves · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Identified string resonances
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Frequency / Hz
Horizontal axis (x): Harmonic number
Read the exact data as a table
| Harmonic number | Frequency / Hz |
|---|---|
| 2 | 50 |
| 3 | 75 |
| 4 | 100 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A string fixed at both ends is 1.20 m long. Its three lowest observed resonances are 50, 75 and 100 Hz. The 50 Hz pattern has two loops, each between adjacent nodes. The supplied graph relates the identified harmonic number to resonance frequency. Wave speed is unchanged between these observations. Supplied data (Harmonic number; Frequency / Hz): (2, 50), (3, 75), (4, 100).
- (a)
Determine the unobserved fundamental, wave speed and node spacing at 75 Hz.
3 marks - (b)
The tension is quadrupled while length and mass per length remain fixed. Predict the resonances and explain the roles of nodes and antinodes.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Determine the unobserved fundamental, wave speed and node spacing at 75 Hz.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
For both ends fixed, L = nλ/2. Two loops identify 50 Hz as the second harmonic, not the fundamental.
- 2
The fundamental is 50/2 = 25 Hz. Wave speed is 2Lf₁ = 2(1.20)(25) = 60 m s⁻¹.
- 3
At 75 Hz, λ = 60/75 = 0.80 m; adjacent nodes are λ/2 = 0.40 m apart.
(b)
The tension is quadrupled while length and mass per length remain fixed. Predict the resonances and explain the roles of nodes and antinodes.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
String wave speed follows v = √(T/μ), so multiplying tension by four doubles speed and all resonance frequencies.
- 2
At a node, destructive superposition keeps displacement zero; at an antinode the displacement amplitude is maximum.
- 3
The standing pattern arises from incident and reflected waves of equal frequency travelling in opposite directions. It is not a single wave carrying the pattern along the string.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- For both ends fixed, L = nλ/2. Two loops identify 50 Hz as the second harmonic, not the fundamental.
- The fundamental is 50/2 = 25 Hz. Wave speed is 2Lf₁ = 2(1.20)(25) = 60 m s⁻¹.
- At 75 Hz, λ = 60/75 = 0.80 m; adjacent nodes are λ/2 = 0.40 m apart.
- String wave speed follows v = √(T/μ), so multiplying tension by four doubles speed and all resonance frequencies.
- At a node, destructive superposition keeps displacement zero; at an antinode the displacement amplitude is maximum.
- The standing pattern arises from incident and reflected waves of equal frequency travelling in opposite directions. It is not a single wave carrying the pattern along the string.
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